Question

An organic compound ‘A’ (\(\mathrm{C_9H_{10}O}\)) when treated with conc. HI undergoes cleavage to yield compounds ‘B’ and ‘C’. ‘B’ gives yellow precipitate with \(\mathrm{AgNO_3}\) where as ‘C’ tautomerizes to ‘D’. ‘D’ gives positive iodoform test. ‘A’ could be

• A.

  

• B.

  

• C.

  

• D.

  

Hint:

B is an alkyl iodide (AgI is yellow) and C must be an enol that tautomerizes to a methyl carbonyl compound (acetaldehyde). So the ether must be R-O-CH=CH₂ in which the R-O bond (sp³ carbon) cleaves.

Answer 1

The Correct Option is C

All four options are \(\mathrm{C_9H_{10}O}\). HI cleaves an ether so that I^- attacks the sp³ alkyl carbon; an aryl-O or vinyl-O bond is not broken by I^-.
(1) p-CH₃C₆H₄-O-CH=CH₂: no sp³ C-O bond, so it would not cleave to give an alkyl iodide and an enol. Rejected.
(2) C₆H₅-O-CH₂CH=CH₂: gives phenol + allyl iodide; phenol does not tautomerize to a carbonyl compound giving the iodoform test. Rejected.
(3) C₆H₅CH₂-O-CH=CH₂: HI cleaves the sp³ C-O bond to give B = C₆H₅CH₂I (benzyl iodide, gives yellow AgI with AgNO₃) and C = CH₂=CH-OH, which tautomerizes to D = CH₃CHO. CH₃CHO contains the CH₃CO- group and gives a positive iodoform test. Accepted.
(4) C₆H₅-O-CH=CH-CH₃: no sp³ C-O bond; gives phenol and no suitable enol/iodide pair. Rejected.
Answer: (3) C₆H₅CH₂-O-CH=CH₂ (benzyl vinyl ether)